# std::next_permutation

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< cpp‎ | algorithm

Defined in header `<algorithm>` |  |
template< class BidirIt > bool next_permutation( BidirIt first, BidirIt last ); | (1) | (constexpr since C++20)
template< class BidirIt, class Compare > bool next_permutation( BidirIt first, BidirIt last, Compare comp ); | (2) | (constexpr since C++20)

Permutes the range `[`first`,` last`)` into the next permutation. Returns true if such a “next permutation” exists; otherwise transforms the range into the lexicographically first permutation (as if by std::sort) and returns false.

1) The set of all permutations is ordered lexicographically with respect to operator<(until C++20)std::less{}(since C++20).
2) The set of all permutations is ordered lexicographically with respect to comp.

If the type of *first is not Swappable(until C++11)`BidirIt` is not ValueSwappable(since C++11), the behavior is undefined.

### Parameters

first, last | - | the range of elements to permute
comp | - | comparison function object (i.e. an object that satisfies the requirements of Compare) which returns true if the first argument is less than the second. The signature of the comparison function should be equivalent to the following: bool cmp(const Type1& a, const Type2& b); While the signature does not need to have const&, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) `Type1` and `Type2` regardless of value category (thus, Type1& is not allowed, nor is Type1 unless for `Type1` a move is equivalent to a copy(since C++11)). The types Type1 and Type2 must be such that an object of type BidirIt can be dereferenced and then implicitly converted to both of them.
Type requirements
- `BidirIt` must meet the requirements of LegacyBidirectionalIterator.

### Return value

true if the new permutation is lexicographically greater than the old. false if the last permutation was reached and the range was reset to the first permutation.

### Complexity

Given \(\scriptsize N\)N as std::distance(first, last):

1,2) At most \(\scriptsize \frac{N}{2}\)

N
2

swaps.

### Exceptions

Any exceptions thrown from iterator operations or the element swap.

### Possible implementation

    template<class BidirIt>
    bool next_permutation(BidirIt first, BidirIt last)
    {
        auto r_first = std::make_reverse_iterator(last);
        auto r_last = std::make_reverse_iterator(first);
        auto left = std::is_sorted_until(r_first, r_last);

        if (left != r_last)
        {
            auto right = std::upper_bound(r_first, left, *left);
            std::iter_swap(left, right);
        }

        std::reverse(left.base(), last);
        return left != r_last;
    }

### Notes

Averaged over the entire sequence of permutations, typical implementations use about 3 comparisons and 1.5 swaps per call.

Implementations (e.g. MSVC STL) may enable vectorization when the iterator type satisfies LegacyContiguousIterator and swapping its value type calls neither non-trivial special member function nor ADL-found `swap`.

### Example

The following code prints all three permutations of the string "aba".

Run this code

    #include <algorithm>
    #include <iostream>
    #include <string>

    int main()
    {
        std::string s = "aba";

        do
        {
            std::cout << s << '\n';
        }
        while (std::next_permutation(s.begin(), s.end()));

        std::cout << s << '\n';
    }

Output:

    aba
    baa
    aab

### See also

is_permutation (C++11) | determines if a sequence is a permutation of another sequence (function template)
prev_permutation | generates the next smaller lexicographic permutation of a range of elements (function template)
ranges::next_permutation (C++20) | generates the next greater lexicographic permutation of a range of elements (algorithm function object)

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